SCERT AP 7th Class Maths Solutions Pdf Chapter 8 Exponents and Powers Ex 8.1 Textbook Exercise Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 8th Lesson Exponents and Powers Ex 8.1

Questions 1.
Express the following into exponential form:.
(i) 14 × 14 × 14
14 × 14 × 14 = 143

(ii) 25 × 25 × 25 × 25 × 25
25 × 25 × 25 × 25 × 25 = 25

(iii) ab × ab × ab × ab
ab × ab × ab × ab = (ab)4

(iv) 7 × p × p × q
7 × p × p × q = 7 × p2 × q Question 2.
Express the following into expanded form:
(i) 276
276 = 27 × 27 × 27 × 27 × 27 × 27

(ii) 1015
1015 = 101 × 101 × 101 × 101 × 101

(iii) (2b)4
2b4 = 2b × 2b × 2b × 2b

(iv) 3a8
3a8 = 3 × a× a × a × a × a × a × a × a

Question 3.
Express the following In exponential form through prime factorisation: .
(i) 81 81 = 3 × 27
= 3 × 3 × 9
= 3 × 3 × 3 × 3
∴ 81 = 34

(ii) 125 125 = 5 × 25
= 5 × 5 × 5
∴ 125 = 53

(iii) 324 324 = 2 × 162
= 2 × 2 × 81
= 2 × 2 × 3 × 11
= 2 × 2 × 3 × 3 × 9
= 2 × 2 × 3 × 3 × 3 × 3
∴ 324 = 22 × 34

(iv) 1080 1080 = 2 × 540
= 2 × 2 × 270
= 2 × 2 × 2 × 135
= 2 × 2 × 2 × 3 × 45
= 2 × 2 × 2 × 3 × 3 × 15
= 2 × 2 × 2 × 3 × 3 × 3 × 5
∴ 1080 = 23 × 31 × 51 Question 4.
Compute and identify the greater num-ber in the following pairs : 7
(i) 25 or 52
25 = 2 × 2 × 2 × 2 × 2 = 32
52 = 5 × 5 = 25
32 > 25
Therefore, 25 > 52

(ii) 73 or 37
73 = 7 × 7 × 7 = 343
37 = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 2187
2187 > 343
Therefore, 37 > 73

(iii) 23 or 32
23 = 2 × 2 × 2 = 8
32 = 3 × 3 = 9
9 > 8
Therefore, 32 > 23

Question 5.
Expand 33 × 42 and 43 × 32. Are they equal? Justify.
33 × 42 = 3 × 3 × 3 × 4 × 4
= 27 × 16 = 432

43 × 32 = 4 × 4 × 4 × 3 × 3
= 64 × 9 = 576

432 × 576 (They are not equal)
Therefore, 576 > 432
So, 43 × 32 > 33 × 42

Question 6.
Express the following numbers in exponential form with the given base.
(i) 1000, base 10 ∴ 1000 = 10 × 10 × 10 = 103

(ii) 512 base 2 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
∴ 512 = 29

(iii) 243 base 3 243 = 3 × 3 × 3 × 3 × 3
∴ 243 = 35 Question 7.
If a = 2, b = 3 find the value of
(i) aa + bb
Given a = 2, b = 3
aa + bb = 22 + 33
= (2 × 2) + (3 × 3 × 3)
= 4 + 27 = 31
Therefore, if a = 2, b = 3 then aa + bb = 31

(ii) ab + ba
Given a = 2, b = 3
ab + ba = 23 + 32
= (2 × 2 × 2) + (3 × 3).
= 8 + 9 = 17
Therefore, if a = 2, b = 3
then ab + ba = 17

(iii) (a + b)b
Given a = 2, b = 3
(a + b)b = (2 + 3)3 = 53
= 5 × 5 × 5 = 125
Therefore, if a = 2, b == 3
then (a + b)b =125

Question 8.
Write the following in Exponential form:
(i) The speed of light in vacuum is about 30,00,00,000 m/sec.